Respuesta :
Answer:
[tex]h=2.5456\ m[/tex]
Explanation:
Given:
- pressure of air in the duct, [tex]P_1=105\ kPa[/tex]
- temperature of air in the duct, [tex]T_1=37^{\circ}C[/tex]
- diameter of the duct, [tex]d_d=0.06\ m[/tex]
- flow rate through the duct, [tex]\dot V_1=65\ L.s^{-1}=0.065\ m^3.s^{-1}[/tex]
- diameter after reduction, [tex]d_r=0.04\ m[/tex]
- difference in the datum of the two points, [tex]z_2=0.2\ m[/tex]
Using continuity equation:
[tex]\dot V_1=\dot V_2[/tex]
[tex]\therefore \dot V_2=0.065\ m^3.s^{-1}[/tex]
As we know,
[tex]\dot V_2=A_2.v_2[/tex]
[tex]A_2=\pi\times \frac{d_r^2}{4}[/tex]
[tex]A_2=\pi\times \frac{0.04^2}{4}[/tex]
[tex]A_2=\pi\times 10^{-4}\ m^3.s^{-1}[/tex]
Hence:
[tex]v_2=\frac{\dot V}{A_2}[/tex]
[tex]v_2=\frac{0.065}{\pi\times 10^{-4}}[/tex]
[tex]v_2=206.9\ m.s^{-1}[/tex]
And:
[tex]v_1=\frac{\dot V_1}{A_1}[/tex]
[tex]v_1=0.065\div \pi \times \frac{0.06^2}{4}[/tex]
[tex]v_1=22.99\ m.s^{-1}[/tex]
According to Bernoulli's principle we have for an incomprehensible flow:
[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} +z=constt.[/tex]
Now we're concerned about the density of air:
[tex]P.V=m.R.T[/tex]
[tex]\rho=\frac{P}{R.T}[/tex]
[tex]\rho=\frac{105000}{0.167\times 310}[/tex]
[tex]\rho=1.18\ kg.m^{-3}[/tex]
Apply Bernoulli's principle assuming it an in-compressible flow:
[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} +z_1=\frac{P_2}{\rho.g} +\frac{v_2^2}{2g} +z_2[/tex]
[tex]\frac{105000}{1.18\times 9.8} +\frac{22.99^2}{2\times 9.8} +0=\frac{P_2}{1.18\times 9.8} +\frac{206.9^2}{2\times 9.8} +0.2[/tex]
[tex]P_2=80053.04\ Pa[/tex]
Now the difference in pressure:
[tex]\Delta P=105000-80053.04[/tex]
[tex]\Delta P=24946.96\ Pa[/tex]
∴The corresponding difference in the water levels of manometer:
[tex]\Delta P=\rho.g.h[/tex]
[tex]24946.96=1000\times9.8\times h[/tex]
[tex]h=2.5456\ m[/tex]
